1 . Type of isomerism exists between :-

CH3—CH2—CH2—CN and CH 3 CH CH 3

CN

(A) Position (B) Chain (C) Both the above (D) None of these

2 . The number of primary alcohols possible with the formula C4H10O is -

(A) 2 (B) 3 (C) 4 (D) 5

3 . The total number of benzene derivatives having the molecular formula C7H7Cl is -

(A) 2 (B) 3 (C) 4 (D) 5

4 . Which of the following will lead to maximum enolisation :–

(A) CH3COCH3 (B) CH3COCH2CHO

(C) CH 3

O

C CH

O

C CH3

Br

(D) O

5 . C C H

C

HC3

HC3H

COOHHC3

Exhibits :-

(A) Tautomerism (B) Optical isomerism

(C) Geometrical isomerism (D) Geometrical and optical isomerism

6 . Tautomerism is not exhibited by :-

(A) (B) (CH3)2CHNO2

(C) (D) (CH3)3CCHO

7 . Given compound exhibits x geometrical isomers and y optical isomers

The value of x and y respectively are :-

(A) 4 and 4 (B) 2 and 2 (C) 2 and 4 (D) 4 and 2

8 . The total number of cyclic compounds (neglecting stereoisomers) with the molecular formula C5H10 is -

(A) 4 (B) 5 (C) 6 (D) 7

9 . Which of the following exhibits tautomerism ?

(A) (CH3)2NH (B) (CH3)3CNO2 (C) R3CNO2 (D) RCH2NO2

EXERCISE-01 CHECK YOUR GRASP

SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER)

1 0 . The correct structure of trans–2–hexenal is -

(A) CHO (B) CHO

(C) CHO (D) CHO

1 1 . The total number of benzene derivatives having the molecular formula C7H8O is -

(A) 3 (B) 4 (C) 5 (D) 6

1 2 . Which of the following compounds does not exhibit tautomerism

(A) CH3NO2 (B) CH3CH2NO2

(C) C6H5CH=CH–OH (D) CH3CH2OH

1 3 . The total number of benzene derivatives with the molecular formula C6H3Cl3 is -

(A) 2 (B) 3 (C) 4 (D) 5

1 4 . Which of the following compounds is not chiral ?

(A) DCH2CH2CH2Cl (B) CH3CHDCH2Cl (C) CH3CHClCH2D (D) CH3CH2CHDCl

1 5 . The total number of stereoisomers of 2,3–dibromobutane is -

(A) 2 (B) 3 (C) 4 (D) 5

1 6 . In the structure :

CH3

HH

OHBr

CH3

the configurations at the chiral centres are :

(A) 2R, 3R (B) 2S, 3R (C) 2R, 3S (D) 2S, 3S

1 7 . Which of the following compound contains a pseudo-asymmetric carbon atom

(A) CH3CHCHCH2

Br Br Br

(B)CHCH–CH CHCH3 3–

Br OH Br

(C) CH3CHCHCHCH3

OH BrBr

(D) CH3CHCHCHCH3

Br OHCl

1 8 . Consider the following structures (A), (B), (C) and (D) -

(A)

CH3

CH52

Cl Br (B)

CH52

CH3

Cl Br

(C)

CH52

CH3

Cl

Br (D) CH52CH3

Cl

Br

Which of the following statements is not correct

(A) B and C are identical (B) A and B are enantiomers

(C) A and C are enantiomers (D) B and D are enantiomers

1 9 . The interchange of two groups (Br and CH3) at the chiral centre of the projection formula (A) yields the

formula (B), while the interchange of another set of two groups (C2H5 and Cl) of (A) yields the projection

formula (C) -

CH3

CH52

CH52

CH52

CH3

CH3

Cl Cl

Cl

Br

Br

Br

(A) (B) (C)

Which of the following statements is not correct about the structures (A), (B) and (C) -

(A) B and C are identical (B) A and C are enantiomers

(C) B and C are enantiomers (D) A and B are enantiomers

2 0 . How many meso stereoisomers are possible for 2, 3, 4-pentanetriol -

(A) 1 (B) 2 (C) 3 (D) none of these

2 1 . The total number of stereoisomers of the compound

CHCH=CH–CH–CH=CH–CH3 3

OH

is -

(A) 2 (B) 3 (C) 4 (D) 8

2 2 . The total number of aldehydes and ketones with the molecular formula C4H8O is -

(A) 2 (B) 3 (C) 4 (D) 5

2 3 . In which of the following properties do enantiomers differ from each other

(A) Solubility in an achiral solvent (B) Reactivity with an achiral reagent

(C) Melting point (D) Optical rotation

2 4 . (+) - Mandelic acid has a specific rotation of + 158º. What would be the observed specific rotation of a

mixture of 25% (–) -mandelic acid and 75% (+) -mandelidc acid :

(A) + 118.5° (B) –118.5° (C) – 79° (D) + 79°

2 5 . When C6H

4Cl

2 is converted into C

6H

3Cl

3, o-isomer will give m types of C

6H

3Cl

3, and p-isomer will give

q types of C6H

3Cl

3. m, n, q are respectively :

(A) 1, 2, 3 (B) 2, 1, 3 (C) 1, 3, 2 (D) 2, 3, 1

CHECK YOUR GRASP EXERCISE -1ANSWER KEYQ u e . 1 2 3 4 5 6 7 8 9 10 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 19 2 0

An s . B A C D B D B A D B C D B A B B B D C D

Q u e . 2 1 2 2 2 3 24 25

An s . C B B D B

SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS)

1 . Tautomerism is shown by :-

(A) HCN (B)

O O

CH—C CH3 3—CH—2 C—

(C) CH—3 CH—N2

O

O(D) CCl3CHO

2 . In which of the following cases, cis-trans nomenclature can not be used :-

(A) Cl—CH=CH—Cl (B) CH3—CH=CH—CHO

(C) C6H5—N=N—C6H5 (D) CH3—CH=C(Cl)C2H5

3 . Among the following, which are tautomers :-

(A) CH—C2 —CH3

O

and CH=C2 —CH3

O

(B) CH3—CH2—CH=NH and CH3—CH=CH—NH2

(C) CH—3 NO

O and CH3—CH=N—OH (D) glucose and fructose

4 . Consider the following compounds :

H———OH

CH3

CH3

HO———H

(I)

HO———H

CH3

CH3

HO———H

(II)

HO———H

CH3

CH3

H———OH

(III)

Choose the correct statements :

(A) I and III are enantiomers (B) I and II are diastereomers

(C) II and III are diastereomers (D) I, II and III are all optically active

5 . Which will show geometrical isomerism :-

(A) C6H5–CH=NOH (B)

(C) C6H5–N=N–C6H5 (D) H

HC3

H

CH3

6 . Which statement (s) is/are correct for :-

CHO

C H6 5

HO———H

(I)

H———OH

(II)

CHO

C H6 5

HO———H

H———OH

(A) Both are in threo form (B) Both are enantiomers

(C) Both are diastereomers (D) Both are in erythro form

EXERCISE–02 BRAIN TEASERS

7 . Which of the following compounds are chiral and resolvable :-

(A) C6H5N(CH3) (C2H5) (B) [C6H5 N

(CH3) (C2H5) (C3H7)] Cl–

(C) CH3–CH2–CH(CH3) N (CH3) (C2H5) (D)

COOH

COOH

8 . Observe the following structures and pick up the correct option (s) mentioned below :-

CH3

H

H H

Cl

CH2Cl(A)

CH3

H

H H

Cl

Cl(B)

(A) The two are position isomers

(B) None of the two shows optical isomerism

(C) Only A shows optical isomerism

(D) The two are not related to each other regarding isomerism

9 .

H———Cl

CH3

CHCH2 3

Cl —H——

(I)

H———Cl

CH3

CHCH2 3

H —Cl——

(II)

(A) I and II are enantiomers (B) I is 2S, 3S; while II is 2S, 3R

(C) I is 2R, 3R; While II is 2R, 3S (D) I and II are diastereomers

1 0 . Which of the following statements are true regarding following structures :-

H———OH

COOCH3

COOH

H———OH

(a)

H———OH

COOCH3

COOH

H———OH

(b)

H———OH

COOCH3

COOH

HO———H

(c)

(A) A and B are diastereomers (B) A and C are diastereomers

(C) B and C are diastereomers (D) A and B are enantiomers

1 1 . The R and S enantiomers of an optically active compound differ in :-

(A) their reactivity with chiral reagents

(B) their melting points

(C) their optical rotation of plane polarized light

(D) their solubility in achiral reagents

1 2 . Which of the following statements (s) is (are) incorrect :-

YX

(A) X is cis- and Y is trans (B) X is Z and Y is E

(C) X is trans and Y is cis (D) X and Y are diastereomers

1 3 . Which of the following compounds will show geometrical isomerism :-

(A) 2-butene (B) propene (C) 1-phenylpropene (D) 2-methyl-2-butene

1 4 . Tautomerism is exhibited by :-

(A) – –OHCH=CH (B) =OO= (C) O

O

(D) O

O

1 5 . Which of the following notations are correct :

(A) C=CBr

I

F

Cl(Z)

(B) C=CBr

I

Cl

I(E)

(C) C=CHC3 CH3

(Z)H H

(D) C=C

(E)

CHCH CH2 2 3HC3H CH3

1 6 . For which of the following pairs of compounds are the correct notation given :-

(A) N=N

Ph Ph

and N=N

Ph

PhAnti-azobenzene Syn-azobenzene

(B)

HC3C=N

H OH

Syn-acetaldoxime

and

HC3C=N

H

Anti-acetaldoxime

OH

(C)

C=C

HCOH2H

NH2

C=C

H

COH2

H

NH2

and

Trans-o-aminocinnamic acid Cis-o-aminocinnamic acid

(D) C=C

CHCH CH2 2 3HC3H CH3

and C=CCH3

ClCH2

CH3BrH2C

Z-isomer E-isomer

1 7 . Which of the following statements are correct :-

(A) CH3CH2CH2CH2OH and

CH3

CH CHOH3 2CH represent chain isomerism

(B) CH3CH2CH2CH=CH2 and CH3CH2CH = CHCH3 are examples of position

isomerism

(C) C2H5OCH3 and CH3CH2CH2OH represent functional-group isomerism

(D) CH3CH2NH2 and CH3NHCH3 are examples of metamerism

1 8 . Which of the following are optically active :-

(A) H2C=C=CH2 (B)

C=C=C

(C) HN2

HH

NH2

(D)

F

F

F

F

—COH2

Cl

1 9 . An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The

ester formed will be :

(A) optically active mixture (B) pure enantiomer

(C) meso compound (D) racemic mixture

2 0 . Tartaric acid molecule contains two asymmetric carbon atoms. The number of optical isomers of tartaric

acid is :-

(A) 2 (B) 3 (C) 4 (D) 5

Qu e. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

An s . A,B,C B,C,D B,C A,B,C A,B,C,D A,B B,C A,C B,D B,C,D A,C A,B A,C A,C,D A,C

Qu e. 16 17 18 19 20

An s . B,C,D A,B,C,D B,C A B

BR AIN TE ASERS EXERCISE -2ANSWER KEY

TRUE OR FALSE :

1 . Stereo-isomers which are not mirror image of each other are known as diastereoisomers.

2 . In every case, a pair of enantiomers have a mirror-image relationship.

3 . If a compound has an enantiomer it must be chiral.

4 . If a compound has a diastereomer it must be chiral

5 . Any molecule containing a stereocentre must be chiral.

6 . Any chiral compound with a single asymmetric carbon must have a positive optical rotation if the compound

has the R configuration.

FILL IN THE BLANKS :

1 . The possible number of dichloro derivatives of propane are ...................... .

2 . Ethyl benzene is ................... isomer to xylenes.

3 . The compound CHCl = CHCl can show .................... isomerism.

4 . d- and - lactic acid are known as ................ .

5 . Maleic and fumaric acids are a pair of ............... .

MATCH THE COLUMN

1 . Match the column I with column II.

Column-I (react ion) Column-II (stereoisomers)

( A ) CH3—CH CH—CH N—OH ( p ) 2

( B ) ( q ) 4

(C) CH3—CH CH—CH CH—CH CH—CH

3( r ) 6

(D) CH3—CH CH—CH CH—CH CH—Ph ( s ) 8

2 . Match the following compounds of column I with column II.

Column-I (Molecule) Column-II (Proper t y)

( A ) C C CH

Cl

H

Cl( p ) Chiral compound

( B )CH3

CH3H

H( q ) Presence of stereocenter

( C )Br F

Cl

I

( r ) Optically active compound

( D ) C NCH3

Et

OH

••

( s ) Compound containing plane

of symmetry

EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS

3 . Match the column I with column II.

Column-I (react ion) Column-II (stereoisomers)

( A )Cl

& Cl (p) Homologs

( B )

OCH –2 3CH

&

OCH3

CH3

(q) Functional isomers

( C )

OCH3

&

OH

CH3

(r ) Metamer

( D ) CH–CH–3 2 –CH–CH2 3 & (s ) Chain isomers

CH–3 – –CH3CH–CH2 2

ASSERTION & REASON QUESTION :

These questions contains, Statement-I (assertion) and Statement-II (reason).

(A) Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I

(B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I

(C) Statement-I is True, Statement-II is False.

(D) Statement-I is False, Statement-II is True.

1 . Statement-I : Staggered and eclipsed ethane can not be separated.

B e c au s e

Statement-II : Energy barrier between staggered and eclipsed form of ethane is 12.6 kJ/mole.

2 . Statement-I : All double bond containing compounds show geometrical isomerism.

B e c au s e

Statement-II : Alkenes have restricted rotation about the double bond.

3 . Statement-I : Meso-tartaric acid is optically active.

B e c au s e

Statement-II : Optically active molecule is a molecule that cannot be superimposed on its mirror image.

4 . Statement-I : Cyclohexanone exhibits keto-enol tautomerism.

B e c au s e

Statement-II : In cyclohexanone, one form contains the keto group (C=O) whi le other contains enolic

group (–C=C–OH).

5 . Statement-I : Staggered form is less stable than the eclipsed form.

B e c au s e

Statement-II : The conformation in which the bond pairs of two central atoms are very far from one

another is called staggered from.

6 . Statement-I : Trans-isomers are more stable than cis-isomer.

B e c au s e

Statement-II : The cis-isomer is the one in which two similar groups are on the same side of double

bond.

7 . Statement-I : Propadiene is optically inactive.

B e c au s e

Statement-II : Propadiene has a plane of symmetry.

COMPREHENSION BASED QUESTIONS :

Comprehension # 1

Geometrical isomerism is a kind of stereoisomerism which is present in the compounds containing a double

bond (C=C, C=N, N=N) and arise due to the restricted or frozen rotation about the double bond. The atoms

or groups attached to the doubly bonded carbons must be different. In aldoximes, the isomer is named as

syn if hydrogen and hydroxyl groups are on the same side of C=N bond and if these are on opposite sides,

the isomer is named as anti. In ketoximes, the prefixes syn and anti indicate which group of ketoxime is syn

or anti to hydroxyl group.

1 . Which of the following does not show geometrical isomerism ?

(A) 1,2-Dichloropent-1-ene (B) 1,3-Dichloropent-2-ene

(C) 1,1-Dichloropent-1-ene (D) 1,4-Dichloropent-2-ene

2 . On treating with NH2OH, which can form two products ?

(A) Acetaldehyde (B) Acetone

(C) Formaldehyde (D) Benzophenone

3 . Number of stereoisomers of the compound 2-chloro-4-methylhex-2-ene is/are

(A) 1 (B) 2 (C) 4 (D) 16

4 . The correct structure of trans-2-hexenal is -

(A) CHO

(B)

CHO

(C) CHO (D) CHO

5 . The geometrical isomerism is shown by :

(A)

CH2

(B)

CH2

(C)

CHCl

(D)

CHCl

Comprehension # 2

The optical isomers rotate the plane of plane-polarised light. A sp3-hybridised carbon atom attached to four

different atoms or groups is called an asymmetrical centre or chiral centre. Chiral molecules do not possess

any of the elements of symmetry. A chiral molecule cannot be superimposed on its mirror image. These

stereoisomers are called enantiomers. Molecules having a plane of symmetry or centre of symmetry are

superimposable on their mirror images and are achiral. The stereoisomers that are not mirror images of

each other are called diastereomers. A mesoisomer has a plane of symmetry and is optically inactive due to

internal compensation.

1 . Which of the following has a meso isomer also ?

(A) 2-Chlorobutane (B) 2-3, Dichlorobutane

(C) 2, 3-Dichloropentane (D) 2-Hydroxypentanoic acid

2 . Which of the following compounds is not chiral ?

(A) DCH2CH2CH2Cl (B) CH3CH2CHDCl

(C) CH3CHDCH2Cl (D) CH3CHClCH2D

3 . The total number of acylic isomers including the stereoisomers (geometrical and optical) possible with the

molecular formula C4H7Cl is :

(A) 12 (B) 11 (C) 10 (D) 9

4 . Which among the following compounds will be dissymmetric but not asymmetric :

(A)

COOH

H—C—OH

H—C—OH

CH3

(B)

OH

CH—C—COOH3

C H2 5

(C)

COOH

H—C—Br

H—C—Br

COOH

(D)

CH3

C=C=C

CH3

H H

5 . Two isomeric alkenes A and B have molecular formula C5H9Cl. On adding H2, A gives optically inactive

compound while B gives chiral compound. The two isomers are :

(A) A is 3-Chlorpoent-1-ene and B is 1-chloropent-2-ene

(B) A is 2-Chloro-3-methylbut-2-ene while B is 1-Chloro-3-methylbut-1-ene

(C) A is 3-Chloropent-2-ene and B is 2-Chloropent-2-ene

(D) A is 4-Chloropent-2-ene and B is 4-Chloropent-1-ene

Comprehension # 3

Different spatial arrangements of the atoms that result from rotation about a single bond are conformers.

n-Butane has four conformers eclipsed, fully eclipsed, gauche and anti. The stability order of these conformers

are as follows :

Anti > gauche > Partial eclipsed > Fully eclipsed

Although anti is more stable than gauche but in some cases gauche is more stable than anti.

1 . Which one of the following is most stable conformer :

(A)

ClH3C

H

H

ClCH3

(B)

Cl

H CH3

CH3

H

Cl

(C)

Cl

H CH3

H Cl

CH3

(D)

Cl

H CH3

HCl

HH3C

Tr ue / Fa l se

1. T 2. T 3. T 4. F 5. T 6. F

Fi l l i n the B lanks

1. four 2. chain 3. geometrical 4. enantiomer

5. geometrical

Match the Co lumn

1. (A) q ; B p ; (C) r ; (D) s 2. (A) p, q, r ; (B) q, s ; (C) p, q, r ; (D) q, s

3. (A) p ; (B) r ; (C) q ; (D) r, s

Asser t i on - Reason Ques t ions

1. A 2. D 3. D 4. B 5. D

6. B 7. A

Comprehens ion Based Que st ions

Comprehens ion #1 : 1. (C) 2. (A) 3. (C) 4. (B) 5. (D)

Comprehens ion #2 : 1. (B) 2. (A) 3. (C) 4. (D) 5. (C)

Comprehens ion #3 : 1. (A) 2. (C) 3. (D)

MISCELLANEOUS TYPE QUESTION EXERCISE -3ANSWER KEY

2 . Which one of the following is the most stable conformer ?

(A) H OH

HO H

CH3

CH3

(B) H OH

OH

H CH3

CH3

(C) H OH

HCH3

CH3

OH

(D) H

HOH

CH3

OHCH3

3 . Number of possible conformers of n-butane is :

(A) 2 (B) 4 (C) 6 (D) infinite

SUBJECTIVE QUESTIONS :

1 . How many isomers are there corresponding to the formula C4H

10O ?

2 . If the bonds in dichloro benzene, C6H

4Cl

2, were localized between specific carbon atoms, how may isomers

of this compound would exist ? How many isomers actually exists.

3 . Which of the following compounds can exist as geometric isomers ?

CH2Cl

2, CH

2Cl—CH

2Cl, CHBr = CHCl, CH

2Cl—CH

2Br.

4 . How will you distinguish between Maleic acid and Fumaric acid ?

5 . Why does cyclopentene not exhibit geometric isomerism though it has a double bond.

6 . Why does 2-butene exhibit cis-trans isomerism but 2-yne does not ?

7 . Which of the following pairs show tautomerism.

(a)

H —C—NMe3 2C

OCH3 and

HC—C—NMe3 2

OMe+

(b)

O

N

H

and

HO

N

8 . Write structural isomer of C6H

14. What is relation between them ?

9 . Arrange the following in the order of their enolic content :

O

O

I

O

OII

O

OIII

O

IV

1 0 . Which of the following does/do not exhibits tautomerism.

(i) CH—C—3 CH3

O

, (ii) CH—C—3 CH—C—OC H2 2 5

O O

(iii) CH=CH—2 C—H

O

(iv) —OH

EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE

1 . There are 7—1-butanol, 2-butanol, 2-methyl-1-propanol, 2-methyl-2-propanol, diethyl ether, methyl propyl

ether, and methylisopropyl ether.

2 . If the bonds were localized, there would be 4 isomers ; actually there are only 3 of the following, the first two

are identical, because the bods are not localized.

Cl

Cl

Cl

Cl

Cl

Cl

Cl

Cl3 . Only CHBr = CHCl can exist as geometric isomers :

C CBr

H

Cl

H and C C

Br

H Cl

H

In CH2Cl— CH2Cl and CH2Cl—CH2Br, the carbon atoms are connected by a single bond about which the

groups can rotate relatively freely. Thus any conformation of the halogen atoms may be converted into any

other simply by rotation about the single bond. In CH2Cl2, the configuration of the molecule is tetrahedral and

all interchanges of atoms yield exactly equivalent configurations.

4 . Maleic acid forms an anhydride where as fumaric acid does not.

CH

CH

CO2H

CO2H

H

C

O

C

O

O

H

H

H CO2H

CO2H H

H

no anhydride.

5 .

H H

This is cis form. Two H atoms on the same side. To get trans, ring must be twisted.

Double bond becomes severely twisted-destabilized. Effective overlap of P orbitals is missing, so does not exist.

6 . The Pz orbitals forming -bonds and the empty Pz orbital of the carbon with +ve charge are parallel. So the

electrons may be delocalized. The +ve charge is effectively spread out over two carbons; delocalized.

CH

HC

C+H

-orbital

empty orbital

H

CH2=CH— 2C H

2C H

—CH= CH

In n-propyl cation, + I effect of R increases the stability.

In allyl + M effect increases the stability. But + M effect in allyl cation is more effective. So allyl > propyl.

A group with + M effect stabilizes cation; destabilizes anion.

7 . Lone pair - p conjugation between fluorine and carbon will be more effective than between chlorine and

carbon.

9. III > IV > I > II

CONCEPTUAL SUBJECTIVE EXERCISE EXERCISE -4 (A)ANSWER KEY

SUBJECTIVE QUESTIONS :

1 . What is relation between (a), (b), (c) ?

(a)

NHHN

O

ON

H

O

(b) NN

OH

OHNHO

(c)

NHHN

O

ON

H

O

–

+ +

+––

2 . Which side is favoured at equilibrium, provide quantitative explanation :

O

O O

O

2

a

H O

pK 16

O

O O

O

pK =13a

+ HO–

3 . (+) 2-butanol has specific rotation of + 13.9° when measured in pure form. A sample of 2-butanol was

found to have an optical rotation of –3°. What is the stereomeric composition of this mixture ?

4 . N-methylethenamine as such does not show any stereoisomerism but one of its resonance form exhibit

stereoisomerism. Explain.

5 . Assign Cahn-ingold prelog priorites to the following sets of substituents :

( i ) —H, —Br, —CH2CH

3, —CH

2CH

2OH (ii) —COOH, —COOCH

3, —CH

2OH, —OH

(ii i) —CN, —CH2NH

2, —CH

2NHCH

3, —NH

2(iv) —Br, —CH

2Br, —Cl, —CH

2Cl

6 . Identify whether the stereogenic centre is present or not :

(i) 2-Cyclo penten-1-ol (ii) 3-cyclo penten-1-ol (iii) 2-bromopentane (iii) 3-bromopentane

7 . Discuss the optical activity of tertiary amines of the type R1R

2R

3N :

8 . Draw the enantiomer of the following structure :

H

Cl

C

CH3

C H2 5

9 . 2,4-Hexadiene has three geometrical isomers. Draw their structures.

1 0 . Assign R and S configuration to the chiral carbons in the following :

(i) HO— —H—

CH3

H— —OH—

CH3

(ii) H— —OH—

H— —OH—

CH3

CH3

EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE

1. a & b are tautomers and a & c are resonating structures.

3. Let x is the % of (+) 2-butanol.

13.9 x – 13.9 (100 – x) = – 300.

x = 39.2, % of d form = 39.2, % of form = 60.8.

4.

CH =CH — NH—CH2 3

:

CH —CH = NH—CH2 3

:

(this shows Geometrical isomerism)

5. (a) —Br > —CH2CH

2OH > —CH

2CH

3 >— H, (b) —OH > —COOCH

3 >—COOH >—CH

2OH

(c) —NH2 >—CN >—CH

2NHCH

3 >—CH

2NH

2(d) —Br >—Cl >—CH

2Br >—CH

2Cl

6. (i)

H OH

Hydroxyl bearing carbon is stereogenic centre., (ii)

H OH

4=3 3=4

2=55=2 It has no stereogenic centre.

(iii) CH—C CH3 3—CH —2 2—CH

Br

H

bromine bearing carbon is stereogenic centre.

(vi) CH— CH CH3 3—CHCH— —2 2

Br

It has no stereogenic centre.

7 . Ter tiary amines have pyramideal geometry with sp3-hybridization at nitrogen.

It should be a chiral molecule (assuming lone pair to be a substituent). Thus, N

R1 R3

R2

••

tertiary amines exist as racemic mixture but they cannot be resolved.

This is due to the reason that the energy difference between the isomer is very small (25 kJ mol–1). Hence,

rapid nitrogen or amine inversion takes palce.

R1

R2

R3

N

••

sp2

N

R3R2

R1

• •

sp2

-hybridized

R1

R2

R3

N

••

sp2

Planar transition stateEnantiomers

Tertiary amine N-oxide has four group hence nitrogen inversion is not possible, thus tertiary amine -N-oxide

can be resolved.

8 .

H

Cl C

CH3

C H2 5

10. (i) 2R, 3R (ii) 2S, 3R

BRAIN STORMING SUBJECTIVE EXERCISE EXERCISE -4 (B)ANSWER KEY

EXERCISE–05(A) PREVIOUS YEARS QUESTIONS

1 . Recemic mixture is formed by mixing two- [A IEEE-2002 ]

(1) isomeric compounds (2) chiral compounds

(3) meso compounds (4) enantiomers with chiral carbon

2 . Geometrical isomerism is not shown by- [A IEEE-2002 ]

(1) 1,1–dichloro–1–pentene (2) 1,2–dichloro–1–pentene

(3) 1,3–dichloro–2–pentene (4) 1,4–dichloro–2–pentene

3 . Following types of compounds I and II [A IEEE-2002 ]

(I) CH3CH=CHCH

3

(II) CH3 CH OH

CHCH2 3

, are studied

in terms of isomerism in-

(1) chain isomerism (2) position isomerism

(3) conformers (4) stereo isomerism

4 . Among the following four structures I to IV [A IEEE-2003 ]

CH2 5 CH CH3 7

CH3

(I)

CH2 5CH

CH3

(II)

CH52 C

O

H C

H

H

+

(III)

CH2 5CH

CH3

(IV)

CH2 5

It is true that-

(1) All four are chiral compounds

(2) only I and II are chiral compounds

(3) only III is a chiral compound

(4) only II and IV are chiral compounds

5 . Which of the following will have a meso-isomer also- [A IEEE-2004 ]

(1) 2–chlorobutane (2) 2,3–dichlorobutane

(3) 2,3–dichloropentene (4) 2–hydroxy propanoic acid

6 . Amongst the following compounds, the optically active alkane having lowest molecular mass is

(1) CH3 CH2C CH [AIEEE-2004 ]

(2) CH3 CH2 CH

CH3

CH3

(3) CH3 C

H

CH2 5

(4) CH3 CH2 CH2 CH3

7 . Which of following compounds is not chiral [A IEEE-2005 ]

(1) 1–chloropentane

(2) 2–chloropentane

(3) 1–chloro–2–methyl pentane

(4) 3–chloro–2–methyl pentane

8 . Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is-

(1) 2-methyl pentane [A IEEE-2005 ]

(2) 2,2–dimethyl butane

(3) 2,3–dimethyl butane

(4) n-hexane

9 . Which types of isomerism is shown by 2,3–dichloro butane- [AIEEE-2005]

(1) structural (2) geometric

(3) optical (4) diastereo

1 0 . Increasing order of stabi li t y among the three main conformat ions (i .e. Eclipse, Ant i, Gauche) of

2-fluoroethanol is [A IEEE-2006 ]

(1) Gauche, Eclipse, Anti

(2) Eclipse, Anti, Gauche

(3) Anti, Gauche, Eclipse

(4) Eclipse, Gauche, Anti

1 1 . Which one of the following conformations of cyclohexane is chiral ? [AIEEE - 2007]

(1) Twist boat (2) Rigid

(3) Chair (4) Boat

1 2 . Which of the following molecules is expected to rotated the plane of plane-polarised light ?

(1) H

CHO

CH OH2

HO(2)

SH[AIEEE - 2007]

(3) H H

H N2 NH2

Ph Ph(4)

COOH

HH N2

H

1 3 . The absolute configuration of

HOC2

HO

COH2

OHHH

is [AIEEE - 2008]

(1) S, S (2) R, R (3) R, S (4) S, R

1 4 . D glu cos e and D glu cos e are [AIEEE - 2008]

(1) conformers (2) epimers

(3) anomers (4) enantiomers

1 5 . The alkene that exhibits geometrical isomerism is :- [AIEEE - 2009]

(1) 2–butene (2) 2–methyl–2–butene

(3) Propene (4) 2–methyl propene

1 6 . The number of stereoisomers possible for a compound of the molecular formula

CH3–CH = CH–CH(OH)–Me is :- [AIEEE - 2009]

(1) 4 (2) 6 (3) 3 (4) 2

1 7 . Out of the following, the alkene that exhibits optical isomerism is :- [A IEEE-2010 ]

(1) 2-methyl-2-pentene (2) 3-methyl-2-pentene

(3) 4-methyl-1-pentene (4) 3-methyl-1-pentene

1 8 . Identify the compound that exhibits tautomerism :- [A IEEE-2011 ]

(1) 2-Pentanone (2) Phenol

(3) 2-Butene (4) Lactic acid

1 9 . How many chiral compounds are possible on monochlorination of 2–methyl butane ? [A IEEE-2012 ]

(1) 6 (2) 8 (3) 2 (4) 4

2 0 . Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of mono

substituted alkyl halide ? [A IEEE-2012 ]

(1) Neohexane (2) Tertiary butyl chloride

(3) Neopentane (4) Isohexane

2 1 . How many cyclic structures are possible for C4H6 :- [AIEEE-2012(Online) ]

(1) 3 (2) 5 (3) 4 (4) 6

2 2 . Maleic acid and fumaric acids are :- [A IEEE -2012 (On l i ne ) ]

(1) Tautomers (2) Chain isomers

(3) Geometrical isomers (4) Functional isomers

PREVIOUS YEAR QUESTIONS ANSWER KEY EXERCISE-5(A)

Q u e. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

A n s 4 1 4 2 2 3 1 3 3 2 1 1 2 3 1

Q u e . 16 17 18 19 20 21 22

A n s 1 4 1 4 3 3 3

MCQ's WITH ONE CORRECT ANSWER

1 . The enolic form of acetone contains : [ I IT - 90 ]

(A) 9 bonds, 1 bond and 2 lone pairs (B) 8 bonds, 2 bonds and 2 lone pairs

(C) 10 bonds, 1 bonds and 1 lone pair (D) 9 bonds, 2 bond and 1 lone pair

2 . An organic molecule necessarity shows optical activitiy if it : [ I I T - 93 ]

(A) Contains asymmetric carbon atoms (B) is non polar

(C) is non superimposable on its mirror image (D) is superimposable on its mirror image

3 . The compound which is not isomeric with diethyl ether is : [ I I T - 93 ]

(A) butan–1–ol (B) butanone

(C) 2–methyl propan–2–ol (D) n–propyl methyl ether

4 . Ordinary light can be converted into plane polarized light with the help of a : [ I I T - 93 ]

(A) Nickel prism (B) Nicol prism (C) Diffraction grating (D) Quartz cell

5 . The structure shows : [ I IT -95 ]

C=CH

C

HC3

HC3H

COOHCH3

(A) Geometrical isomerism (B) Optical isomerism

(C) Geometrical & optical isomerism (D) tautomerism

6 . How many optically active stereoisomers are possible for butane –2,3–diol : [ I IT -97 ]

(A) 1 (B) 2 (C) 3 (D) 4

7 . Isomers which can be interconverted through rotation around of single bond are -

[ I IT -97 ]

(A) Conformers (B) Diastereomers (C) Enantiomers (D) Positional isomers

8 . The number of possible enantiomeric pairs that can be produced during monochlorination of

2–methyl butane is : [ I IT -97 ]

(A) 2 (B) 3 (C) 4 (D) 1

9 . Tautomerism is not exhibited by : [ I IT-98]

(A) CH=CH–OH(B) OO (C) O

O

(D) O

O

1 0 . Rotation of polarised light can be measured by : [ I IT -98 ]

(A) Monometer (B) Galvanometer (C) Polarimeter (D) Viscometer

1 1 . The optically active tartaric acid is named as D–(+) tartaric acid because it has a positive : [ I IT-99]

(A) optical rotation and is derived from D–glucose

(B) pH in an organic solvent

(C) optical rotation and is derived from D–(+)– glyceraldehyde

(D) optical rotation only when substituted by deuterium

EXERCISE–05(B) PREVIOUS YEARS QUESTIONS

1 2 . The enol form of acetone, after treatment with D2O gives [ I IT -99 ]

(A) CH–C=CH3 2

OD

(B) CD–C–CD3 3

O

(C) CH–C=CH2 2D

OH

(D) C =C–CD2 3D

OH

1 3 . Which of the following compound will exhibits geometrical isomerism : [ I I T - 2 000 ]

(A) 1–phenyl–2–butene (B) 3–phenyl–1–butene

(C) 2–phenyl–1–butene (D) 1, 1–diphenyl–1–propene

1 4 . The number of isomers for the compound with molecular formula C2BrCIFI is : [ I I T - 2000 ]

(A) 3 (B) 4 (C) 5 (D) 6

1 5 . Which of the following exhibits stereoisomerism– [ I IT - 2000 ]

(A) 2–Methylbutene–1 (B) 3–Methylbutyne–1

(C) 3–Methylbutanoic acid (D) 2–Methylbutanoic acid

1 6 . Hydrogen of the following compound in the presence of poisoned palladium catalyst gives : [IIT-2002]

Me HH

MeHMe

(A) optically active compound (B) an optically inactive compound

(C) a racemic mixture (D) a diastereomeric mixture

1 7 . Which of the following has the lowest dipole moment :

[ I I T - 2002 ]

(A) C=CCH3

H

CH3

H(B) CH3–CC–CH3 (C) CH3CH2CCH (D) CH2=CH–CCH

1 8 . If C2 in below compound is rotated by 120º angle in anticlockwise direction along C2–C3, which of the

following form will be produced : [ I I T - 2004 ]

CH3

HHCH3

H

1

32

4

H

(A) Partial eclipsed (B) Per fectly eclipsed (C) Per fectly staggered (D) Gauche conformation

1 9 . CH–CH–CH3 2–CH3

CH3

Cl /hv2 N no. of isomers Fractional distillation F, (N) and (F) are :

[ I I T - 2006 ]

(A) 6, 4 (B) 4, 4 (C) 6, 6 (D) 3, 3

2 0 . The number of structural isomers of C6H14 is : [ I I T - 2007 ]

(A) 3 (B) 4 (C) 5 (D) 6

2 1 . The number of stereoisomers obtained by bromination of trans–2–butene is : [ I I T - 2007 ]

(A) 1 (B) 2 (C) 3 (D) 4

2 2 . Statement-I : Molecules that are not superimposable on their mirror images are chiral

B e c au s e

Statement-II : All chiral molecules have chiral centres. [ I I T - 2007 ]

(A) Statement-1 is True, Statement-2 is True ; Statement-2 is a correct explanation for Statement-1

(B) Statement-1 is True, Statement-2 is True ; Statement-2 is NOT a correct explanation for Statement-1

(C) Statement-1 is True, Statement-2 is False.

(D) Statement-1 is False, Statement-2 is True.

2 3 . The correct statement(s) concerning the structures E, F and G is (are) [ I I T - 2008 ]

H C3

H C3

O

CH3

(E )

H C3

H C3

OH

CH3

(F )

H C3

H C3

CH3

OH(G )

(A) E, F and G are resonance structures

(B) E, F and E, G are tautomers

(C) F and G are geometrical isomers

(D) F and G are diastereomers

2 4 . The correct statement(s) about the compound given below is (are) :- [ I I T - 2008 ]

Cl H

H C3CH3

Cl H

(A) The compound is optically active

(B) The compound possesses centre of symmetry

(C) The compound possesses plane of symmetry

(D) The compound possesses axis of symmetry

2 5 . The correct statement(s) about the compound H3C(HO)HC – CH = CH – CH(OH)CH

3 (X) is (are) :

(A) The total number of stereoisomers possible for X is 6 [ I I T - 2009 ]

(B) The total number of diastereomers possible for X is 3

(C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for

X is 4

(D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2

2 6 . The bond energy (in kcal mol–1) of a C–C single bond is approximately : [ I I T - 2010 ]

(A) 1 (B) 10

(C) 100 (D) 1000

2 7 . In the Newman projection for 2,2–dimethylbutane [ I IT - 2010 ]

XCH3

HH

H C3

YX and Y can respectively be –

(A) H and H (B) H and C2H

5(C) C

2H

5 and H (D) CH

3 and CH

3

2 8 . Amongst the given option, the compound(s) in which all the atoms are in one plane in all the possible

conformations (if any), is (are) - [ I IT-2011]

(A)

HC – C

H

H C2 CH2

(B)

H

CH2

H – C C – C (C) H2C = C = O (D) H

2C = C = CH

2

2 9 . Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ? [ I I T - 2012 ]

H HH

Cl CH3

HO HO Cl

HO H HO

CH3 Cl CH3

Cl OH

OH

M N O

H

CH3 CH3

OH H

HO HO

Cl Cl

H H

H HO

P Q

(A) M and N are non-mirror image stereoisomers (B) M and O are identical

(C) M and P are enantiomers (D) M and Q are identical

P & Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O, On heating P forms the cyclic

anhydride.

Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from

S, T and U. [JEE ADVANCED-2013]

COOH COOH COOH

H H HO

H HO H

OH OH H

OH H OH

COOH COOH COOH

(S) (T) (U)

3 0 . Compounds formed from P and Q are respectively

(A) Optically active S and optically active pair (T, U)

(B) Optically inactive S and optically inactive pair (T, U)

(C) Optically active pair (T, U) and optically active S

(D) Optically inactive pair (T, U) and optically inactive S

PREVIOUS YEARS QUESTIONS EXERCISE -5 (B)

1. (A) 2. (C) 3. (B) 4. (B) 5. (B) 6. (B) 7. (A) 8. (A)

9. (B) 10. (C) 11. (C) 12. (A) 13. (A) 14. (D) 15. (D) 16. (B)

17. (B) 18. (D) 19. (B) 20. (C) 21. (A) 22. (C) 23. (B, C, D)

24. (A, D) 25. (A, D) 26. (C) 27. (B, D) 28. (B, C) 29. (A, B, C) 30. (B)

ANSWER KEY